(a) Find: ∫ (e^x)/(√(5-4 e^x - e^2 x)) dx

Let $\displaystyle \mathrm{e}^{\mathrm{x}}=\mathrm{t}$, so that $\displaystyle \mathrm{e}^{\mathrm{x}} \mathrm{dx}=\mathrm{dt}$. Then,

$$\begin{aligned} & \mathrm{I}=\int \frac{\mathrm{dt}}{\sqrt{$\displaystyle 5$-$\displaystyle 4$ \mathrm{t}-\mathrm{t}^{$\displaystyle 2$}}}=\int \frac{d \mathrm{t}}{\sqrt{$\displaystyle 3$^{$\displaystyle 2$}-(\mathrm{t}+$\displaystyle 2$)^{$\displaystyle 2$}}} & =\sin ^{-$\displaystyle 1$}\left(\frac{\mathrm{t}+$\displaystyle 2$}{$\displaystyle 3$}\right)+\mathrm{C} & =\sin ^{-$\displaystyle 1$}\left(\frac{\mathrm{e}^{\mathrm{x}}+$\displaystyle 2$}{$\displaystyle 3$}\right)+\mathrm{C} \end{aligned}

$\displaystyle \mathrm{I}=\int_{0}^{\pi / 2} \sqrt{\sin \mathrm{x}} \cdot\left(1-\sin ^{2} \mathrm{x}\right)^{2} \cos \mathrm{x} d \mathrm{x}$

Put $\displaystyle \sin \mathrm{x}=\mathrm{t}^{2}$

$$\begin{gathered} \left.=\int_{$\displaystyle 0$}^{$\displaystyle 1$} \sqrt{\mathrm{t}}\left($\displaystyle 1$-\mathrm{t}^{$\displaystyle 2$}\right)^{$\displaystyle 2$} \mathrm{dt}=\int_{$\displaystyle 0$}^{$\displaystyle 1$} \sqrt{\mathrm{t}}+\mathrm{t}^{$\displaystyle 9$ / $\displaystyle 2$}-$\displaystyle 2$ \mathrm{t}^{$\displaystyle 5$ / $\displaystyle 2$}\right) \mathrm{dt} \left.=\frac{$\displaystyle 2$ \mathrm{t}^{$\displaystyle 3$ / $\displaystyle 2$}}{$\displaystyle 3$}+\frac{$\displaystyle 2$ \mathrm{t}^{$\displaystyle 11$ / $\displaystyle 2$}}{$\displaystyle 11$}-\frac{$\displaystyle 4$ \mathrm{t}^{$\displaystyle 7$ / $\displaystyle 2$}}{$\displaystyle 7$}\right] \int_{$\displaystyle 0$}^{$\displaystyle 1$}=\frac{2}{3}+\frac{2}{11}-\frac{4}{7}=\frac{64}{231} \end{gathered}